Question

A record store owner finds that 20% of customers entering her store make a purchase. One morning 180 people, who can be regarded as a random sample of all customers, enter the store.
a. What is the mean of the distribution of the sample proportion of customers making a purchase?
b) What is the variance of the sample proportion?
c) What is the standard error of the sample proportion?
d) What is the probability that the sample proportion is less than 0.15?

Answer 1

Answer:

a) 0.2

b) 0.0009

c) 0.0298

d) 0.0465 = 4.65% probability that the sample proportion is less than 0.15.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

20% of customers entering her store make a purchase.

This means that $p = 0.2$

180 people

This means that $n = 180$

a. What is the mean of the distribution of the sample proportion of customers making a purchase?

By the Central Limit Theorem, $\mu = p = 0.2$.

b) What is the variance of the sample proportion?

The standard deviation is:

$s = \sqrt{\frac{0.2*0.8}{180}} = 0.0298$

Variance is the square of the standard deviation, so:

$s^2 = (0.0298)^2 = 0.0009$

c) What is the standard error of the sample proportion?

As found in the previous item, 0.0298.

d) What is the probability that the sample proportion is less than 0.15?

This is the p-value of Z when X = 0.15. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.15 - 0.20}{0.0298}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that the sample proportion is less than 0.15.